package DynamicProgrammingPackage;

public class longestCommonSubsequence_ {

    // 1. dp[i][j]表示范围在[0,i - 1]的text1和范围在[0,j - 1]的text2的最长公共子序列的长度
    // 2. 递推公式: if(text1.charAt(i - 1) == text2.charAt(j - 1)) 则 dp[i][j] = dp[i - 1][j - 1] + 1
    // else dp[i][j] = Math.max(dp[i - 1][j],dp[i][j - 1])因为一旦子序列不同, 就取得之前遍历过的子序列的长度
    // 3. 初始化dp[0][j] = dp[i][j] = 1
    // 4. 遍历顺序: 从上至下, 从左到右
    public int longestCommonSubsequence(String text1, String text2) {
        int[][] dp = new int[text1.length() + 1][text2.length() + 1];
        for (int i = 0; i < text1.length(); i++) {
            dp[i][0] = 0;
        }
        for (int i = 0; i < text2.length(); i++) {
            dp[0][i] = 0;
        }
        int maxValue = 0;
        for (int i = 1; i <= text1.length(); i++) {
            for (int j = 1; j <= text2.length(); j++) {
                if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
                maxValue = Math.max(maxValue, dp[i][j]);
            }
        }
        return maxValue;
    }
}
